Adding two functions / multiplying two functions yields wrong answer python
I've made a simple example to try and add two functions (for now, want multiply in the end but neither of them work so I decided to try adding).
def addFun(fun1,fun2):
fun = lambda x: fun1(x) + fun2(x)
return fun
d = [12,10]
B = 5
fun1 = lambda d,L: (d-L)**2
result1 = lambda L: 0
for i in range(0,len(d)):
func = lambda L: fun1(float(d[i]),L)
result1 = addFun(result1,func)
print(func(1))
print(result1(1))
The print functions are there to check whether it's working or not (so far it's not. So ideally what should happen is (d-L)^2 for d = 12 is (12-L)^2. I add that to result1 which is now (12-L)^2. Then on the next loop it becomes (10-L)^2. Now the addFun function should add these two together, so the new result1 should be
(12-L)^2 + (10-L)^2. Then I substitute L as 1, answer should be 11^2 + 9^2 = 202. However this is not the case. When I run this through, the resulting answer is 162. Worst thing is the + sign doesn't commute; when I put in 10 first and 12 second, the answer becomes 242.
I've had this working for simpler equations, such as adding x^2 with x^3 e.t.c. But does anyone know why this isn't working here?
python function addition
asked Nov 13 '18 at 16:07
ZWangZWang
665
add a comment |
I've made a simple example to try and add two functions (for now, want multiply in the end but neither of them work so I decided to try adding).
def addFun(fun1,fun2):
fun = lambda x: fun1(x) + fun2(x)
return fun
d = [12,10]
B = 5
fun1 = lambda d,L: (d-L)**2
result1 = lambda L: 0
for i in range(0,len(d)):
func = lambda L: fun1(float(d[i]),L)
result1 = addFun(result1,func)
print(func(1))
print(result1(1))
The print functions are there to check whether it's working or not (so far it's not. So ideally what should happen is (d-L)^2 for d = 12 is (12-L)^2. I add that to result1 which is now (12-L)^2. Then on the next loop it becomes (10-L)^2. Now the addFun function should add these two together, so the new result1 should be
(12-L)^2 + (10-L)^2. Then I substitute L as 1, answer should be 11^2 + 9^2 = 202. However this is not the case. When I run this through, the resulting answer is 162. Worst thing is the + sign doesn't commute; when I put in 10 first and 12 second, the answer becomes 242.
I've had this working for simpler equations, such as adding x^2 with x^3 e.t.c. But does anyone know why this isn't working here?
python function addition
asked Nov 13 '18 at 16:07
ZWangZWang
665
don't dox lambda functions
– wim
Nov 13 '18 at 16:10
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
1
dandiare being looked up by name in yourfunclambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value ofi, the one it had in the final iteration of the loop. Usual solution is to write it aslambda L, i=i:so that you capture the current value ofias a default parameter.
– jasonharper
Nov 13 '18 at 16:37
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58
add a comment |
I've made a simple example to try and add two functions (for now, want multiply in the end but neither of them work so I decided to try adding).
def addFun(fun1,fun2):
fun = lambda x: fun1(x) + fun2(x)
return fun
d = [12,10]
B = 5
fun1 = lambda d,L: (d-L)**2
result1 = lambda L: 0
for i in range(0,len(d)):
func = lambda L: fun1(float(d[i]),L)
result1 = addFun(result1,func)
print(func(1))
print(result1(1))
The print functions are there to check whether it's working or not (so far it's not. So ideally what should happen is (d-L)^2 for d = 12 is (12-L)^2. I add that to result1 which is now (12-L)^2. Then on the next loop it becomes (10-L)^2. Now the addFun function should add these two together, so the new result1 should be
(12-L)^2 + (10-L)^2. Then I substitute L as 1, answer should be 11^2 + 9^2 = 202. However this is not the case. When I run this through, the resulting answer is 162. Worst thing is the + sign doesn't commute; when I put in 10 first and 12 second, the answer becomes 242.
I've had this working for simpler equations, such as adding x^2 with x^3 e.t.c. But does anyone know why this isn't working here?
python function addition
asked Nov 13 '18 at 16:07
ZWangZWang
665
I've made a simple example to try and add two functions (for now, want multiply in the end but neither of them work so I decided to try adding).
def addFun(fun1,fun2):
fun = lambda x: fun1(x) + fun2(x)
return fun
d = [12,10]
B = 5
fun1 = lambda d,L: (d-L)**2
result1 = lambda L: 0
for i in range(0,len(d)):
func = lambda L: fun1(float(d[i]),L)
result1 = addFun(result1,func)
print(func(1))
print(result1(1))
The print functions are there to check whether it's working or not (so far it's not. So ideally what should happen is (d-L)^2 for d = 12 is (12-L)^2. I add that to result1 which is now (12-L)^2. Then on the next loop it becomes (10-L)^2. Now the addFun function should add these two together, so the new result1 should be
(12-L)^2 + (10-L)^2. Then I substitute L as 1, answer should be 11^2 + 9^2 = 202. However this is not the case. When I run this through, the resulting answer is 162. Worst thing is the + sign doesn't commute; when I put in 10 first and 12 second, the answer becomes 242.
I've had this working for simpler equations, such as adding x^2 with x^3 e.t.c. But does anyone know why this isn't working here?
python function addition
python function addition
asked Nov 13 '18 at 16:07
ZWangZWang
665
asked Nov 13 '18 at 16:07
ZWangZWang
665
asked Nov 13 '18 at 16:07
ZWangZWang
665
asked Nov 13 '18 at 16:07
ZWangZWang
665
asked Nov 13 '18 at 16:07
ZWangZWang
665
665
don't dox lambda functions
– wim
Nov 13 '18 at 16:10
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
1
dandiare being looked up by name in yourfunclambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value ofi, the one it had in the final iteration of the loop. Usual solution is to write it aslambda L, i=i:so that you capture the current value ofias a default parameter.
– jasonharper
Nov 13 '18 at 16:37
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58
add a comment |
don't dox lambda functions
– wim
Nov 13 '18 at 16:10
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
1
dandiare being looked up by name in yourfunclambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value ofi, the one it had in the final iteration of the loop. Usual solution is to write it aslambda L, i=i:so that you capture the current value ofias a default parameter.
– jasonharper
Nov 13 '18 at 16:37
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58
don't dox lambda functions
– wim
Nov 13 '18 at 16:10
don't dox lambda functions
– wim
Nov 13 '18 at 16:10
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
1
1
d and i are being looked up by name in your func lambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value of i, the one it had in the final iteration of the loop. Usual solution is to write it as lambda L, i=i: so that you capture the current value of i as a default parameter.– jasonharper
Nov 13 '18 at 16:37
d and i are being looked up by name in your func lambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value of i, the one it had in the final iteration of the loop. Usual solution is to write it as lambda L, i=i: so that you capture the current value of i as a default parameter.– jasonharper
Nov 13 '18 at 16:37
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58
add a comment |
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don't dox lambda functions
– wim
Nov 13 '18 at 16:10
@wim why not? What's wrong with doing that in python
– ZWang
Nov 13 '18 at 16:13
1
dandiare being looked up by name in yourfunclambda - when you finally execute the whole thing at the end, all of the individual lambdas will see the same value ofi, the one it had in the final iteration of the loop. Usual solution is to write it aslambda L, i=i:so that you capture the current value ofias a default parameter.– jasonharper
Nov 13 '18 at 16:37
@jasonharper Yeah thats worked thank you very much!
– ZWang
Nov 13 '18 at 19:58